Question

If the function \( f(x) \) is given by \[ f(x) = \begin{cases} \frac{\tan(a(x-1))}{\frac{x-1}{x}}, & tif0<x<1 
\frac{x^3-125}{x^2 - 25} , & \text{if } 1 \leq x \leq 4 
\frac{b^x - 1}{x}, & \text{if } x>4 \end{cases} \] is continuous in its domain, then find \( 6a + 9b^4 \).

• A. 284
• B. 261
• C. 214
• D. 317 \bigskip

Hint:

When solving continuity problems, always check the limits from both sides at the points of discontinuity and equate them to ensure continuity.

Answer 1

The Correct Option is A

The given function \( f(x) \) is: \[ f(x) = \begin{cases} \frac{\tan(a(x-1))}{\frac{x-1}{x}}, & \text{if } 0<x<1
\frac{x^3-125}{x^2 - 25} , & \text{if } 1 \leq x \leq 4
\frac{b^x - 1}{x}, & \text{if } x>4 \end{cases} \] To ensure continuity, we must check at \( x = 1 \) and \( x = 4 \). --- Step 1: Continuity at \( x = 1 \) For continuity at \( x = 1 \), we need: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \] # Left-hand limit (\( x \to 1^- \)): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{\tan(a(x-1))}{\frac{x-1}{x}} \] For small \( x - 1 \), using the approximation \( \tan u \approx u \): \[ \tan(a(x-1)) \approx a(x-1) \] Thus, \[ \frac{\tan(a(x-1))}{\frac{x-1}{x}} \approx \frac{a(x-1)}{\frac{x-1}{x}} = a \cdot x \] Taking \( x \to 1 \): \[ \lim_{x \to 1^-} f(x) = a(1) = a \] # Right-hand limit and \( f(1) \): \[ f(1) = \frac{1^3 - 125}{1^2 - 25} = \frac{1 - 125}{1 - 25} = \frac{-124}{-24} = \frac{31}{6} \] Equating limits: \[ a = \frac{31}{6} \] --- Step 2: Continuity at \( x = 4 \) For continuity at \( x = 4 \): \[ \lim_{x \to 4^-} f(x) = \lim_{x \to 4^+} f(x) = f(4) \] # Left-hand limit (\( x \to 4^- \)) and \( f(4) \): \[ f(4) = \frac{4^3 - 125}{4^2 - 25} = \frac{64 - 125}{16 - 25} = \frac{-61}{-9} = \frac{61}{9} \] # Right-hand limit (\( x \to 4^+ \)): \[ \lim_{x \to 4^+} \frac{b^x - 1}{x} = \frac{b^4 - 1}{4} \] Equating: \[ \frac{b^4 - 1}{4} = \frac{61}{9} \] Solving for \( b \): \[ b^4 - 1 = \frac{244}{9} \] \[ b^4 = \frac{253}{9} \] \[ b = \left(\frac{253}{9}\right)^{\frac{1}{4}} \] --- Step 3: Compute \( 6a + 9b^4 \) \[ 6a + 9b^4 = 6 \times \frac{31}{6} + 9 \times \frac{253}{9} \] \[ = 31 + 253 = 284 \] --- Final Answer: \[ \boxed{284} \] \bigskip